One can continue like this to compute $\displaystyle X:= \sum_{n=0}^\infty \frac{n^3}{2^n}$. We have $\displaystyle X = 2X-X = \sum_{n=0}^\infty \frac{3n^2+3n+1}{2^n} = 3S+3T+R = 26$. Sums with larger powers can be computed in the same way.

summed first in the columns, then in the rows. It doesn't matter which direction you sum first, because all sums in both directions converge absolutely. So we can sum first in the rows, and then in the columns. Each row-sum is just a geometric series; the $m$-th row-sum is $r_m=\sum_{n=m}^{\infty} 2^{-n}=2^{1-m}$. Now summing up all rows, we have $\sum r_m = 2$.

Let me show you a slightly different approach; this approach is very powerful, and can be used to compute values for a large number of series.

Let's think of this in terms of power series. You noticed that you can write
$$
\sum_{n=0}^{\infty}\frac{n^2}{2^n}=\sum_{n=0}^{\infty}n^2\left(\frac{1}{2}\right)^n;
$$
so, let's consider the power series
$$
f(x)=\sum_{n=0}^{\infty}n^2x^n.
$$
If we can find a simpler expression for the function $f(x)$, and if $\frac{1}{2}$ lies within its interval of convergence, then your series is exactly $f(\frac{1}{2})$.

Now, we note that
$$
f(x)=\underbrace{0}_{n=0}+\sum_{n=1}^{\infty}n^2x^n=x\sum_{n=1}^{\infty}n^2x^{n-1}.
$$
But, notice that $\int n^2x^{n-1}\,dx=nx^n$; so,
$$\tag{1}
\sum_{n=1}^{\infty}n^2 x^{n-1}=\frac{d}{dx}\left[\sum_{n=0}^{\infty}nx^n\right].
$$
Now, we write
$$\tag{2}
\sum_{n=0}^{\infty}nx^n=\underbrace{0}_{n=0}+x\sum_{n=1}^{\infty}nx^{n-1}=x\frac{d}{dx}\left[\sum_{n=0}^{\infty}x^n\right].
$$
But, this last is a geometric series; so, as long as $\lvert x\rvert<1$,
$$
\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}.
$$
Plugging this back in to (2), we find that for $\lvert x\rvert<1$,
$$
\sum_{n=0}^{\infty}nx^n=x\frac{d}{dx}\left[\frac{1}{1-x}\right]=x\cdot\frac{1}{(1-x)^2}=\frac{x}{(1-x)^2}.
$$
But, plugging this back in to (1), we find that
$$
\sum_{n=1}^{\infty}n^2x^{n-1}=\frac{d}{dx}\left[\frac{x}{(1-x)^2}\right]=\frac{x+1}{(1-x)^3}
$$
So, finally,
$$
f(x)=x\cdot\frac{x+1}{(1-x)^3}=\frac{x(x+1)}{(1-x)^3}.
$$
Plugging in $x=\frac{1}{2}$, we find
$$
\sum_{n=0}^{\infty}\frac{n^2}{2^n}=f\left(\frac{1}{2}\right)=6.
$$